Page 67 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
16:18
Brown.cls
Brown˙C02
a
x
C BEAMS b B = 0 49
A = C/L B = −C/L
y
y
FIGURE 2.24 Free-body-diagram.
U.S. Customary SI/Metric
Example 1. Determine the reactions at the Example 1. Determine the reactions at the
ends of a simply-supported beam of length (L) ends of a simply-supported beam of length (L)
with a concentrated couple (C) acting at an with a concentrated couple (C) acting at an
intermediate point, where intermediate point, where
C = 15 ft · kip = 15,000 ft · lb C = 20 kN · m = 20,000 N · m
1
1
L = 12 ft, a = 4 ft, b = 8ft L = 4m, a = 1 m, b = 2 m
2 2
solution solution
Step 1. From Fig. 2.24 calculate the pin reac- Step 1. From Fig. 2.24 calculate the pin reac-
tions (B x and B y ) at the right end of the beam. tions (B x and B y ) at the right end of the beam.
As there are no forces acting on the beam, As there are no forces acting on the beam,
B x = 0 B x = 0
and the vertical reaction (B y ) is and the vertical reaction (B y ) is
C −15,000 ft · lb C −20,000 N · m
B y =− = B y =− =
L 12 ft L 4m
=−1,250 lb =−5,000 N
Step 2. From Fig. 2.24 calculate the roller Step 2. From Fig. 2.24 calculate the roller
reaction (A y ) at the left end of the beam. reaction (A y ) at the left end of the beam.
C 15,000 ft · lb C 20,000 N · m
A y = = A y = =
L 12 ft L 4m
= 1,250 lb = 5,000 N

Shear Force and Bending Moment Distributions. For the simply-supported beam with
a concentrated couple (C) at an intermediate point, shown in Fig. 2.25, which has the
balanced free-body-diagram shown in Fig. 2.26, the shear force (V ) distribution is shown in
Fig. 2.27.

a b
C
A B

L
FIGURE 2.25 Concentrated couple at intermediate point.